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Hmm...Looking at that problem, if "trapezium" means trapezoid, which I'm assuming it does...well, it's the average of 7 and 10 times the perpindicular distance between EF and CB. However...in order to get that distance, I believe you need to use a right triangle...but you're missing some important measurements. You have to find EB, and then square it and subtract 1.5 squared, then find the square root of that...that's your perpindicular distance. It's finding EB that's the problem. If you just had the height of the figure...ech...

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yes but... wouldnt EB be 5cm? since AB is 5cm, and the sides of the roof are equalatiral triangles...

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Since AB = EB, I thought you could simply use Pythagoras with the height of the trapezium being the square root of (5^2 - 1,5^2) but then you get the no-good answer of 4,77 cm

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Ah. I missed the equilateral bit.

Anyways, if EB=5, then just make a perpindicular segement from E to a point we can call X. Triangle XEB is right. EB=5, BX=1.5. 25-2.25=22.75. So it's 8.5 times the square root of 22.75. Do that with a calculator and you have the answer.

That has to be correct, given the numbers we were given...If the triangles are equilateral, the BX has to equal 1.5...and since EB=5, we know root 22.75 is correct. So the first answer is the square root of 22.75, the second answer is root 22.75 times 8.5. Root 22.75 is about 4.77, so it's 4.77 times 8.5...that makes it 40.545...That's the answer.

5. My mind goes blank if I'm just reading math and I'm too lazy to write it out on paper.

So... err...

*nods head and pretends CI & Saizou & PST are correct*
Originally Posted by Saizou
If you'd like I could post the BASTARD QUESTION FROM HELL that was on last year's finals. I think that three people of nearly fifty got it right.

Sure! Post it so I can feel stupid. =D

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Wow a whole year of sleeping in math class which I used to be one of the best at can really affect you later on. And I mean really.

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Originally Posted by coolerimmortal
Ah. I missed the equilateral bit.

Anyways, if EB=5, then just make a perpindicular segement from E to a point we can call X. Triangle XEB is right. EB=5, BX=1.5. 25-2.25=22.75. So it's 8.5 times the square root of 22.75. Do that with a calculator and you have the answer.

That has to be correct, given the numbers we were given...If the triangles are equilateral, the BX has to equal 1.5...and since EB=5, we know root 22.75 is correct. So the first answer is the square root of 22.75, the second answer is root 22.75 times 8.5. Root 22.75 is about 4.77, so it's 4.77 times 8.5...that makes it 40.545...That's the answer.
wait... you've lost me. where's X? and just to be sure... is the perpendicular distance the distance between EF and BC, aka the slope of the roof?

oh wait sack that i totally get it now- thanks so much!! ^^ does anyone get b) ? ... the size of the angle EMN? at first i thought it was 90 degrees, but then i thought no, cos the roof's rapezium is a trapezium... meaning that the angle would be acute...

i've gotten so far find out the perpendicular distance of the equilateral triangles so i know EM, and MN is 10cm... and i dunno where to go from there. i dont think i can use the cosine or sine rule...

and also...
six numbers are: 17, 23, 28, 25, a, b
where a and b are two positive integers between 14 and 25.
the 6 numbers have a mean of 22 and a standard deviation of 4.
find the values of a and b
you are advised to use the statistical functions on your calculator, and trial and improvement
Last edited by Quiraikotsu; 05-31-2006 at 03:39 AM.

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I've created a monster....

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A and B...They add to make 39. It can be:
15, 24
16, 23
17, 22
18, 21
19, 20

I forgot how standard deviation works; you'll just have to figure out which one of those is correct.

For the angle measure...Inverse cosine(1.5root3/9). Plug that into a calculator, and you get...wait for it...30.

The angle is 30 degrees.

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